Re: Linear algebra. A proof

I think you are working at this too hard. We know that m(AB) = (mA)B = A(mB) since scalar multiplication commutes with the A and B operations. So you have

(ab)(AB) = (abA)B = (aAb)B = (aA)(bB)

-Dan

Re: Linear algebra. A proof

Thank you. I see, and a proof is that short? Can you explain it more, so I can understand that proof, please.

Re: Linear algebra. A proof

Quote:

Originally Posted by

**ThinkingNumbers24** Thank you. I see, and a proof is that short? Can you explain it more, so I can understand that proof, please.

No real proof is needed. By definition we know that, for a scalar n and an operator N that scalar multiplication is defined as

nN = Nn...Any operator commutes with a scalar. So that gives (ab)AB = (abA)B. And the rest is simply moving b to the other side of A.

It occurs to me that you might actually have to prove aA = Aa from another list of assumptions?

-Dan

Re: Linear algebra. A proof

Thank you again. I just have that I want to prove

(ab)AB = (aA)(bB), in which a, b E R and matrix multiplication AB are defined.

I thought that above can be proven without by definitions.

Well, indeed, I think more, I believe that I will understand your answers.